Let's say you created a cannon that can shoot small black holes and you shoot it at some star.

  1. Would the star just turn into a black hole silently? Or rather first destabilize and produce a last dying blast like supernovas do?

  2. For high enough speed, the black hole would just pop through the star and fly away. How fast would it need to go for that?

(Asking for a friend, I don't actually have a black hole cannon.)

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    $\begingroup$ This will depend a lot on the size of the small black hole. How small were you thinking? A solar mass? A lunar mass? A few kg? $\endgroup$
    – TimRias
    Commented Dec 5, 2023 at 12:48
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    $\begingroup$ @TimRias for sure big enough to not evaporate before it hits the star. And small enough to realistically be able to shoot it. But let's say you can choose whatever size you think would be most practical for a cannon. $\endgroup$ Commented Dec 5, 2023 at 15:44
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    $\begingroup$ Black holes aren't a special kind of matter. It's just a lot of matter that holds together, without a counteracting force to expand it, but it doesn't cause a viral effect. $\endgroup$
    – Therac
    Commented Dec 5, 2023 at 20:34
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    $\begingroup$ Related: physics.stackexchange.com/q/506845/123208 $\endgroup$
    – PM 2Ring
    Commented Dec 5, 2023 at 20:48
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    $\begingroup$ @MarkRansom Hawking radiation. See astronomy.stackexchange.com/questions/tagged/hawking-radiation especially astronomy.stackexchange.com/q/31635/16685 $\endgroup$
    – PM 2Ring
    Commented Dec 6, 2023 at 23:29

4 Answers 4


How much matter can a small black hole absorb? This determines the answer to your questions.

A simple approximation is that moving black holes absorb everything they pass through out to a radius $kR_S$ where $k>1$ determines how widely they can grab passing matter. For a pretty fast hole this is likely less than 2. $R_S$ is the Schwarschild radius of the hole. So if a hole shoots through matter of density $\rho$ for a distance $d$, it will absorb about $\Delta M = \pi \rho d k^2 R_S^2$ kilograms. If a black hole starts with velocity $v$ it will have momentum $v_1M$ at the start, and at the end the momentum $v_2 (M+\Delta M)$ must be the same, leading to an exit velocity $v_2 = v_1 (M/(M+\Delta M))$. For that exit velocity to be above solar escape velocity $v_{esc}=\sqrt{2GM_\odot/R_\odot}$ (about 600 km/s) we must have $v_1/v_{esc}>1+\frac{\Delta M}{M}$, or $$v_1/v_{esc} > 1+ 8 \pi G^2 \rho R_\odot k^2 M/c^4.$$ This tells us a few things. Just dropping a black hole into the sun will make $v_1/v_{esc}=1$, and it will stay. $8\pi G^2 \rho R_\odot k^2/c^4\approx 5.4\cdot 10^{-41}$: the amount of matter a black hole will accumulate when passing through is really small.

What about a black hole just sitting still inside the sun? It will attract nearby matter, but as it gets compressed it will also heat up and the accretion will tend to push away the matter. The limit when the infalling matter is in equilibrium with the pressure is roughly $(1/12)\dot{M}c^2 = 3.4\cdot 10^4 (M/M_\odot)L_\odot$, or $\dot{M}=1.5\cdot10^{32}(M/M_\odot)$ kg/s (assuming thin disk, pure ionized hydrogen and a lot of other stuff which can be debated, especially for big holes). The solar mass is $M_\odot=2\cdot 10^{30}$.

So what this tells us is that if the black hole has a small mass, up to an Earth mass, the amount of absorbed matter is minuscule: there is a hot pinpoint of accretion inside the sun's core, but nothing much changes. Bigger holes will noticeably heat up the core and no doubt have effects on the total brightness. To actually eat the star quickly it needs to have a mass that gets within a percent of the entire star. (Except at that point the above formula doesn't work that well: the star literally cannot free-fall that fast into the black hole.) At this end the final implosion will be pretty bright since the infalling gas get super-compressed and heated, making it produce a bright blast.

Generally, black hole cannons are surprisingly useless.

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    $\begingroup$ @racoon_lord since "black hole cannons are surprisingly useless" it looks like 'your friend' needs to up his/her game! :) $\endgroup$
    – BradV
    Commented Dec 5, 2023 at 16:47
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    $\begingroup$ Would the same analysis hold for firing a small (mountain-mass) black hole through a starship (assuming known materials like steel and aluminum)? It would just punch a pinhole through the ship and make a little accretion disc of atmosphere for a short time, and then it moves on without causing significant harm? $\endgroup$ Commented Dec 5, 2023 at 17:22
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    $\begingroup$ @DarthPseudonym - Yes, black hole bullets merely puncture things. There is a complication for really small holes (like a mountain mass): they have intense Hawking radiation, and may actually do a lot of damage by irradiating the ship with a few gigawatt of x-rays! $\endgroup$ Commented Dec 5, 2023 at 18:01
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    $\begingroup$ In case anybody is interested, $k=\frac{(e+3)^{3/2}}{(e+1)\sqrt{e-1}}$ with $e=\sqrt{1+8v^2}$ $\endgroup$
    – TimRias
    Commented Dec 5, 2023 at 19:10
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    $\begingroup$ A hot point of accretion could have significant effects, as said heat may trigger a new round of fusion which may become self-sustaining or even runaway. I recall reading a paper about testing for and constraining the amount of micro black holes that might compose dark matter because they could trigger white dwarves to go supernova in ways which could be detected and distinguished. $\endgroup$ Commented Dec 7, 2023 at 10:05

@AndersSandberg's answer surprised me so I wanted to poke around some more, to see what could a determined attacker achieve. So here's what would happen if we put a BH in orbit inside a star so that it can continuously feed.

Let's use our Sun as an example, with densities (raw data):


Source code.

For a given $r$, our orbital speed is $v_o\approx \sqrt{\frac{GM_{below}(r)}{r}}$, and mass eaten per $s$ per $m^2$ is $v_o\cdot density$, which gives:

BH will eat the fastest when orbiting at $9\%R_{Sun}$. Below is a simulation of that. I assumed $k=2$ and a BH mass of $13M_{Jupiter}$ (that's the mass of the heaviest known planet). (BH not to scale.)

We can also try just dropping the BH, so that it passes through the dense core. After trial and error, dropping from $\approx 20\%R_{Sun}$ seems best, but not much better than the orbit:

For trajectories in between these two, eating speeds are similar. But if we don't manage to put the BH on such a narrow trajectory, the eating speed is much worse:

Time to consume 1% of the Sun

So for our highest eating speed of $6\cdot 10^{14} kg/s$ it will take about 1 milion years to eat 1% of the Sun's mass. And we used a BH with mass $13M_{Jupiter}$. Eating speed is proportional to $R_{S}^2$ which itself is proportional to mass. So for 10x smaller BH, it feeds 100x slower, which gets hopelessly slow. That square relation also means that shooting many smaller BHs is not worth it - better to use one but big.

But you may have noticed that our $13M_{Jupiter}$ BH is already above 1% of Sun's mass (which Anders proposed as the point where we're getting close to explosion). So even for this already explosive BH that doesn't need to grow, if it was growing through movement, that would still be extremely slow. If we instead used a Jupyter mass BH, it would need to feed for about 150 mln years before it gets explosive.

Some things that may boost the eating rate:

  • maybe for our speeds ($\approx 0.001c$), $k$ is larger
  • movement and accretion may interact in some way that speeds up accretion
  • for larger and denser stars, eating rate through movement would be higher

Some other things I ignored:

  • as BH flies and pulls the matter in, density behind it will be larger, so it will be dragged gravitationally (very slightly, but for long periods this may be significant)
  • opposite effect: maybe the heated up matter behind BH would propel it
  • after a long time BH will lose speed due to gaining mass + conservation of momentum
  • putting a BH on an orbit inside the star is hard
    • you could shoot 2 BHs from opposite directions to merge, but that requires insanely good aim
    • otherwise you could try some elaborate gravitational dance but that's impractical if you're attacking someone else's star system


To kill a star, you'd need to throw in a BH that's already huge (which is possible but really expensive). Smaller BHs will just result in star poisoning, shortening its lifetime, although it's not clear by how much. And the hosts may try to scoop that BH out (which would be the ultimate game of curling).

If the BH heats up the star a lot without destabilizing it, there could be cases where you want to do it intentionally to your own star, to "overclock" it. Later you can also scoop out that BH to go back to normal.

  • $\begingroup$ @Flip Sondej Perhaps you could answer this similar question: How long could world like the Moon survive with an Earth mass black hole in the center? $\endgroup$ Commented Dec 6, 2023 at 19:05
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    $\begingroup$ Hah, I just wondered whether it's worthwhile to watch the simulation until the process escalates. Only after a few minutes I continued reading and stumbled over the "1 million years". I think the industry just lost a few nerd-man days because a handful of people are still meditating over the simulation of a black hole oscillating through a star core and haven't read on yet, out of fear of missing out ;-). Take my upvote for sabotaging capitalism! $\endgroup$ Commented Dec 6, 2023 at 20:29
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    $\begingroup$ "To kill a star, you'd need to throw in a BH that's already huge […]. Smaller BHs will just result in star poisoning […]". Nah. You wanna use lots of smaller black holes. If you toss merely a single Earth mass of individually fishing-ship-massed black holes into the Sun, they should finish evaporating in its core and overcome its gravitational binding energy within a couple minutes. $\endgroup$
    – Will Chen
    Commented Dec 6, 2023 at 21:12
  • $\begingroup$ @WillChen Huh, interesting. Yeah, I guess if you can time it just right so they all explode in the core and you have enough, you may rip the star apart. But how much is enough? IIRC black hole in its last second of life emits energy of about 1 mln megatons of TNT. But Sun is quite massive too. We'd need to calculate it. Note that the production of black holes is likely very expensive, so if it requires millions of BHs, that may be not worth the effort. $\endgroup$ Commented Dec 6, 2023 at 21:41
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    $\begingroup$ @Peter-ReinstateMonica Hahahah <3 Yeah, I also had to clip that gif, because 1 mln year gif wouldn't fit on my laptop. $\endgroup$ Commented Dec 6, 2023 at 21:54

The existing answers haven't really addressed your point 2, so I'll cover that here.

The possibly-surprising thing is that it's basically impossible to shoot a black hole at a star such that it won't go straight through. The reason is that if you're shooting it from outside the star system then it's on a hyperbolic orbit, or at best a parabolic one. In other words, even at the slowest possible speed it's falling into the system from very far away, getting faster and faster as it reaches the star, so that when it gets to the star it's picked up exactly enough speed to escape again. The only way to avoid that would be through gravitational slingshot type interactions with planets, but it will be very difficult to set those up so that the black hole gets captured. (Or by friction as it passes through the star. But as Filip Sondej mentions that's very small for a black hole of with enough mass to actually eat the star - it will barely even slow down.)

Even if your black hole gun is located inside the system you still have a similar problem. You can fire the black hole onto an elliptical orbit that passes through the star, but it's still going to pass through it and continue on its orbit, spending only a small fraction of each orbit inside the star. The orbit will slowly decay until it's entirely inside the star, but that will take a very long time.

  • $\begingroup$ Perhaps you could aim at a planet, creating a larger, slower black hole before reaching the star? $\endgroup$ Commented Dec 7, 2023 at 8:50
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    $\begingroup$ If a star doesn't slow it down, it won't even notice passing through the planet. Would the planet notice? physics.stackexchange.com/questions/568743/… $\endgroup$ Commented Dec 7, 2023 at 21:17

All of the answers here deal with the black hole eating matter as if nothing else happens in the process.

This is plain wrong.

Black holes (well, it is the space neighboring them) are the brightest objects in the astronomy.

Black holes convert between 5% and 40% of the infalling mass into light.

This is how a stellar black hole can outshine an orbiting star, capturing just a small part of its stellar wind. This is also how a theoreticized quazi-stars could exist.

In short, a black hole eating thru a star would create an immense explosion, comparable to supernovas (at least some of them are powered by similar mechanism), possibly completely destroying the star rather quickly.

In the process, only a small part of the stellar mass will sink into the black hole. Most of it will be blown away.

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    $\begingroup$ As the video in the article you linked says, those numbers are the maximum possible efficiency of matter infalling on a spiral course while being slowed by an accretion disk for a nonmoving and rotating black hole respectively, with the light being generated by the collisions in the accretion disk. It's not that all matter falling into a black hole in any possible way begins emitting light. And Anders' answer does specifically mention the radiation pressure from the build-up of an accretion disk eventually balancing out gravitational infall. $\endgroup$
    – Idran
    Commented Dec 8, 2023 at 14:11

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